## College Algebra 7th Edition

$x=0$ or $x=-2$
We solve by factoring: $x^{5}+8x^{2}=0$ $x^{2}(x^{3}+8)=0$ $x^{2}(x+2)(x^{2}-2x+4)=0$ $x^{2}=0$ or $x+2=0$ or $x^{2}-2x+4=0$ $x=0$ or $x=-2$ (The last equation $x^{2}-2x+4=0$ has no real solutions because the discriminant is negative.)